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About Tensor Calculus |
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| Apr5-10, 05:14 AM | #1 |
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About Tensor Calculus
Dear Friends
I have two questions to do about Tensor Calculus: 1) Is there any program to calculate Christoffel Symbols, Riemann and Ricci Tensors and everything about Tensor Calculus (Free or paid)? 2) When in an exercise anyone asks to use the Euclidean metric or Riemann metric, what is the difference? Is it only the presence of € (indicator)? Thank you for your help. Good day. |
| Apr5-10, 10:33 AM | #2 |
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1. Yes. http://grtensor.phy.queensu.ca/
2. I don't know what € is, but a Riemannian metric is any metric with signature (n,0), and a Euclidean metric is a special Riemannian metric of the form ds˛ = dx1˛ + dx2˛ + ... + dxn˛. |
| Apr5-10, 12:08 PM | #3 |
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| Apr5-10, 12:28 PM | #4 |
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About Tensor Calculus
I'm not sure if 'Euclidean metric' has a conventionally accepted frame invariant meaning. I think its just a term used to describe things which look like the Pythagorean formula in some coordinate system. Of course any such coordinate system will be orthonormal since
g(∂i, ∂j) = (dx1˛ + dx2˛ + ... + dxn˛)(∂i, ∂j) = δij. |
| Apr5-10, 12:57 PM | #5 |
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Okay, thanks. So if a metric has Pythagorean form, it's Euclidean, and the form guarrantees that an orthonormal coordinate system is meant.
Since the word tensor is used in this context to mean a tensor (or tensor field) with respect to the tangent space (or if tensor field, the tangent bundle) of a manifold, presumably a Euclidean metric tensor is a frame invariant object that can have coordinate representations besides the diagonal matrix [itex]I[/itex] with entries [itex]\delta_{ij}[/itex] that is its coordinate representation in an orthonormal system. And since the metric can be worked out from the metric tensor, I'm guessing a Euclidean metric can also have non-Pythagorean representations, and that it's the existence of a Pythagorean representation that makes a metric Euclidean, rather than the metric happening to be in Pythagorean form. Is that right? |
| Apr5-10, 02:51 PM | #6 |
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Recognitions:
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For Euclidean, how about a Riemannian manifold with signature X and all components of the curvature tensor zero erverywhere?
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| Apr5-10, 03:26 PM | #7 |
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Thanks for your help.
Till the next. Cheers. |
| Apr5-10, 04:03 PM | #8 |
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| Apr5-10, 04:09 PM | #9 |
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You mean using ∂i for the tangent vector ∂/∂xi?
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| Apr5-10, 04:42 PM | #10 |
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| Apr5-10, 04:56 PM | #11 |
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Since in special relativity, we say that the metric is locally Minkowski, I think it would be consistent to say that the metric on a Riemannian manifold is locally Euclidean. But still, as I said, I don't know if there is a conventional used definition. |
| Apr5-10, 06:08 PM | #12 |
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Ah, okay. Sounds like I was closer to the mark in #3 and #5 than #10.
Is this more or less what you mean by "define it in this way"? A metric field is a function-valued field (in the sense scalar, vector, tensor, etc. field), or perhaps a field whose value at each point is defined as some kind of coordinate-independent equivalence class of functions... Its value at some point, in some coordinate system might be given by, for example, [itex]ds^2 = dr^2 + r^2 \; d \theta^2[/itex]. The metric (i.e. distance function which is the value of the metric field) is said to be Euclidean at this point iff there exists a coordinate system (namely an orthonormal one) in which the coordinates of the metric tensor give the metric the form of Pythagoras's formula. Iff for some n-manifold, there exists a single, global coordinate system according to which the metric everywhere has Pythagorean form, then the manifold is the Euclidean space En. Iff for each point in the underlying set of the manifold, there exists a coordinate system (not necessarily global, i.e. not necessarily the same coordinate system as for all other points) according to which the metric is Euclidean at that point, then the manifold is Riemannian. |
| Apr5-10, 06:23 PM | #13 |
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For software, I use maxima and ctensor. It does everything I need, and it's free and open source.
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| Apr6-10, 05:01 AM | #14 |
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[tex]\hat{e}_{(\mu)} = \partial_{\mu}[/tex] and [tex]\hat{\theta}^{(\mu)} = d^{\mu} \ ,[/tex] as bases in a manner I found too hard to follow. I would be interested in a presentation where the bases dxu are presented as true infinitesimals, and --- I'm not sure what the ∂u would be. |
| Apr6-10, 06:59 AM | #15 |
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[tex] dx_1 \otimes dx_1 + dx_2 \otimes dx_2 + ... + dx_n \otimes dx_n [/tex] and the dxu are coordinate 1-forms. So (dx1˛ + dx2˛ + ... + dxn˛)(∂i, ∂j) would be [tex] (dx_1 \otimes dx_1 + dx_2 \otimes dx_2 + ... + dx_n \otimes dx_n) (\partial_i, \partial_j) = \sum_{\alpha} dx_{\alpha}(\partial_i)dx_{\alpha}(\partial_j) = \sum_{\alpha} \delta_{\alpha i} \delta_{\alpha j} = \delta_{ij} [/tex] If you're interested in GR books that don't use the modern notation of 1-forms and multilinear operators, the only book I can recommend is Landau and Lifshitz's "Classical Theory of Fields". That's probably the only book written in 'old tensor' that is still worth reading. |
| Apr6-10, 07:24 AM | #16 |
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Perhaps this is not such a small wish. You certainly have more modern training than I've had, so I was hoping you might point me in the right direction. |
| Apr6-10, 07:34 AM | #17 |
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Some books (that use modern 'calculus on manifolds') that I like are: "Spacetime, Geometry, Cosmology" by William Burke "Gravitation" by Misner, Thorne and Wheeler "Gauge Fields, Knots and Gravity" by John Baez and Javier Muniain Baez is good for the algebraically oriented, while the other two are good for geometrically oriented readers. |
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